For $\alpha = \sqrt[3]{2} + i$, find the minimal polynomial over $\mathbb{Q}$ and over $\mathbb{Q}[\sqrt[3]{2}]$

Question

I am trying to find a irreducible polynomial over $\mathbb{Q}[x]$ and over $\mathbb{Q}[\sqrt[3]{2}][x]$ that have $\alpha = \sqrt[3]{2} + i$ as zero.

I realized that $p(x) = x^6 + 3x^4 - 4x^3 + 3x^2 + 12x + 5$ is a polynomial in $\mathbb{Q}[x]$ which have $\alpha$ as root. But I failed to prove that $p(x)$ is irreducible over $\mathbb{Q}$.

Answer 1

Another method to show that $p(x)$ is irreducible over $\Bbb Q$ is to show that it is already irreducible over $\Bbb F_7$. This is easier to show.

Answer 2

To prove that the polynomial you found is irreducible over $\mathbb Q$, it s true that you can't use any simple criterion like Eisenstein, but you can work with degrees and dimensions. Let $b=2^{1/3}$. First, prove that $\mathbb Q[a]= \mathbb Q[b,\omega b]=\mathbb Q[b,\omega]$ where $\omega$ is third root of unity. Then, you can see that $[\mathbb Q[b,\omega] : \mathbb Q]=6$ by tower law. But $[\mathbb Q[a],\mathbb Q]= deg irr_{\mathbb Q, a}(x)$= the degree of the minimal polynomial over $\mathbb Q$ with $a$ as its root. However, you ve already found a polynomial of degree 6 with $a$ as a root so it must be irreducible.